Solutions to Problem Set 5
Section 5.1 #8
Equilibrium points: (0,0), (0,30), (10,0), (20,-10)
x
10  2 x  y

J 
30  2 x  2 y 
  2y
By evaluating the Jacobian at each of the equilibrium we find that:
(0,0) is a source, (0,30) is a sink, (10,0) is a saddle and (20,-10) is N/A.
There should be three graphs, one showing the phase portrait near each equilibrium. For
example, near (0,0) all solutions should go away since (0,0) is a source.
Section 5.2 #12
x-nullclines x  0 , y  2  x y-nullclines: y  0 , y  x 2
After you sketch the nullclines and place the correct arrows on them you can split your region
into 4 parts and the axes.
Starting in the top region (call it A) you go off to infinity
Starting in the region to the right (call it B) you either go into A (and hence off to infinity), to the
equilibrium (1,1) or get pulled into the region below it (call it C) and end up going to (2,0)
Starting in C you will go toward (2,0)
The last region (call it D) will have the same behavior as B, you will get pulled into A and hence
to infinity, go to (1,1) or get pulled into region C and go toward (2,0).
If you start on the x-axis solutions go toward (2,0).
If you start on the y-axis solutions go toward infinity.
You can use the information above or your software to sketch the phase portrait.
Section 3.6 #14
d2y
dy
 6  8y  0
2
dt
dt
dy
v
dt
dv
 8 y  6v
dt
1
0
  8  6


  4,2
4
1  x
1
2
1  x
1
  4 ( A  I )V  
    0  V 1    4
  8  2  y 
 
  2 ( A  I )V  
    0  V 2    2
  8  4  y 
 
The oscillator is overdamped since the eigenvalues are real. There is no period. The phase
portrait is a sink with solutions tending to V2 as t goes to infinity. Both the y(t) and v(t) go to
zero, but v(t) starts at zero, and goes negative before going to zero.
Section 3.6 #22
d2y
dy
 6  8y  0
2
dt
dt
 2  6  8  0
(  4)(  2)  0
  4,2
y (t )  k1e  4t  k 2 e  2t
v(t )  4k1e  4t  2k 2 e  2t
k1  1, k 2  2
Graph matches the graphs in problem above.
Section 4.1 #26
d2y
 4 y  2e 2t
2
dt
2  4  0
  4i
y h (t )  k1 cos( 2t )  k 2 sin( 2t )
y p (t )  Ae 2t
A
1
4
y (t )  k1 cos( 2t )  k 2 sin( 2t ) 
1  2t
e
4
1
1
k1   , k 2 
4
4
Solutions will oscillate since the last term will go to 0.
Section 4.3 #10
d2y
 4 y  3 cos 2t
dt 2
2  4  0    2i
y h (t )  k1 cos( 2t )  k 2 sin( 2t )
y p (t )  At cos( 2t )  Bt sin( 2t )
y p ' (t )  A cos( 2t )  2 A sin( 2t )  B sin( 2t )  2 B cos( 2t )
y p ' ' (t )  4 A sin( 2t )  4 B cos( 2t )  4 At cos( 2t )  4 Bt sin( 2t )
A  0, B  3 / 4
y(t )  k1 cos(2t )  k 2 sin( 2t )  3 / 4t sin( 2t )
k1  0, k 2  0
y (t )  3 / 4t sin( 2t )
Section 4.3 #16
d2y
 11y  2 cos 3t
dt 2
2  11  0
   11i
y h (t )  k1 cos( 11t )  k 2 sin( 11t )
y p (t )  A cos(3t )  B sin( 3t )
y p ' (t )  3 A sin( 3t )  3B cos(3t ) , A  1, B  0
y p ' ' (t )  9 A cos(3t )  9 B sin( 3t )
y (t )  k1 cos( 11t )  k 2 sin( 11t )  cos(3t )
Frequency of beat:
11  3
 0.025196 Frequency of fast oscillations:
4
11  3
 .50266
4
These are a factor of 20. The period of the beats is about 40 so the outer curve would go to 40.
Review Chapter 4 #18
d 2 y dy

 2 y  5e  2t
dt 2 dt
2    2  0
(  2)(  1)  0    1,2
y h (t )  k1e t  k 2 e  2t
y p (t )  Ate2t
y p '  2 Ate2t  Ae 2t
y p ' '  4 Ate2t  4 Ae 2t
4 Ate 2t  4 Ae 2t  2 Ate  2t  Ae 2t  2 Ate 2t  5e 2t
 3 A  5  A  5 / 3
5
y (t )  k1e t  k 2 e  2t  te  2t
3
Chapter 4 Review #22
d2y
 3 y  2t  cos 4t
dt 2
2  3  0
   3i
y h (t )  k1 cos( 3t )  k 2 sin( 3t )
y p (t )  At  b  C cos( 4t )  D sin( 4t )
A
2
1
, B  0, C   , D  0
3
13
2
1
y (t )  k1 cos( 3t )  k 2 sin( 3t )  t  cos( 4t )
3 13